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3 years ago

How do the wavelengths of electromagnetic energy absorbed by materials on earth compare?

Physics

1 answer:

Dmitrij [34]3 years ago

8 0

Explanation:

The shorter wavelength electromagnetic waves from sun are absorbed by earth material in form of short wavelength and the radiated wavelength are longer ones. Also higher energy waves are of shorter wavelength and lower energy waves have longer wavelength. So, they are absorbed as short wavelength and radiated back as long wavelength.

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A force of 1.150×103 N pushes a man on a bicycle forward. Air resistance pushes against him with a force of 795 N. If he starts

nalin [4]

Answer:

He has a speed of 16.60m/s after 35.0 meters.

Explanation:

The final velocity can be determined by means of the equations for a Uniformly Accelerated Rectilinear Motion:

$v_{f}^{2} = v_{i}^{2} + 2ad$

$v_{f} = \sqrt{v_{i}^{2} + 2ad}$ (1)

The acceleration can be found by means of Newton's second law:

$\sum F_{net} = ma$

Where $\sum F_{net}$ is the net force, m is the mass and a is the acceleration.

$Fx + Fy = ma$ (2)

All the forces can be easily represented in a free body diagram, as it is shown below.

Forces in the x axis:

$F_{x} = F - F_{air}$ (3)

Forces in the y axis:

$F_{y} = 0$ (4)

Solving for the forces in the x axis:

$F_{x} = F - F_{air}$

Where $F = 1.150x10^{3} N$ and $F_{air} = 795 N$ :

$F_{x} = 1.150x10^{3} N - 795 N$

$F_{x} = 355 N$

Replacing in equation (2) it is gotten:

$Fx + Fy = ma$

$355 N + 0 N = (90.0 Kg)a$

$355 N = (90.0 Kg)a$

$a = \frac{355 N}{90.0Kg}$

$a = \frac{355 Kg.m/s^{2}}{90.0Kg}$

$a = 3.94 m/s^{2}$

So the acceleration for the cyclist is $3.94 m/s^{2}$ , now that the acceleration is known, equation (1) can be used:

$v_{f} = \sqrt{v_{i}^{2} + 2ad}$

However, since he was originally at rest its initial velocity will be zero ( $v_{i} = 0$ ).

$v_{f} = \sqrt{2ad}$

$v_{f} = \sqrt{2(3.94m/s^{2})(35.0m)}$

$v_{f} = 16.60m/s$

He has a speed of 16.60m/s after 35.0 meters

8 0

3 years ago

A double-slit arrangement produces interference fringes for sodium light (λ = 589 nm) that are 0.48° apart. What is the angular

larisa [96]

Answer:

0.38°

Explanation:

$\theta$ = Angle

m = Number

d = Distance

n = Refractive index of liquid = 1.25

a denotes air

l denotes liquid

In the case of double split interferance we have the relation

$m\lambda=dsin\theta$

For air

$m\lambda_a=dsin\theta_a$

For liquid

$m\lambda_l=dsin\theta_l$

Dividing the two equations

$\frac{m\lambda_a}{m\lambda_l}=\frac{dsin\theta_a}{dsin\theta_l}\\\Rightarrow \frac{\lambda_a}{\lambda_l}=\frac{sin\theta_a}{sin\theta_l}$

Wavelength ratio = $n$

$n=\frac{sin\theta_a}{sin\theta_l}\\\Rightarrow \frac{sin0.48}{1.25}=sin\theta_l\\\Rightarrow \theta_l=sin^{-1}\frac{sin0.48}{1.25}\\\Rightarrow \theta_l=0.38^{\circ}$

The angular separation is 0.38°

5 0

3 years ago

Read 2 more answers

The standard measure used to compare sound intensity is the_____

BartSMP [9]

Frequency? Possibly I’m not 100% sure

5 0

3 years ago

An 850-lb force is applied to a 0.15-in. diameter nickel wire having a yield strength of 45,000 psi and a tensile strength of 55

Sindrei [870]

Answer:

a)Yes will deform plastically

b) Will NOT experience necking

Explanation:

Given:

- Applied Force F = 850 lb

- Diameter of wire D = 0.15 in

- Yield Strength Y=45,000 psi

- Ultimate Tensile strength U = 55,000 psi

Find:

a) Whether there will be plastic deformation

b) Whether there will be necking.

Solution:

Assuming a constant Force F, the stress in the wire will be:

stress = F / Area

Area = pi*D^2 / 4

Area = pi*0.15^2 / 4 = 0.0176715 in^2

stress = 850 / 0.0176715

stress = 48,100.16 psi

Yield Strength < Applied stress > Ultimate Tensile strength

45,000 < 48,100 < 55,000

Hence, stress applied is greater than Yield strength beyond which the wire will deform plasticly but insufficient enough to reach UTS responsible for the necking to initiate. Hence, wire deforms plastically but does not experience necking.

6 0

3 years ago

What part causes the disc brake caliper piston to retract when the brakes are released?

Orlov [11]

The part that causes the disc caliper piston to retract when the brakes are released is the square-cut O-ring.
The square cut seal is the most important part of a disc brake caliper, for keeping the brake behind the piston so that when you step on the brake pedal, it releases a pressure that applied to the piston which in return applies the pad to the rotor.

3 0

2 years ago